TITLE: If $A$ is a compact set then so is $A'$?
QUESTION [5 upvotes]: Let $X$ be a metric space with a metric $d$ and let $A$ be a compact subset of $X$.
Show that $A'$ is compact where $A'$ is a derived set of $A$.

I am done $A'$ is closed and bounded.
But we know that if a set is closed and bounded, then the set is not compact, generally.
So I try to prove by using the definition of compactness, but I can't.

REPLY [7 votes]: Hint: Compact sets are closed, and closed subsets of compact sets are compact.